一点处导数的定义公式也适用于导数的导数（二阶导数）

一、题目

已知 $f(x)$ 在 $x=a$ 处二阶可导，请证明：

$$
\lim \limits_{h \rightarrow 0} \frac{f(a+h)+f(a-h)-2 f(a)}{h^{2}}=f^{\prime \prime}(a)
$$

难度评级：

二、解析

$$
\begin{aligned}
& \lim \limits_{h \rightarrow 0} \frac{f(a+h)+f(a-h)-2 f(a)}{h^{2}} \Rightarrow \text{ 洛必达运算 } \Rightarrow \\ \\
& = \lim \limits_{h \rightarrow 0} \frac{f^{\prime}(a+h)-f^{\prime}(a-h)}{2 h} \\ \\
& = \frac{1}{2} \lim \limits_{h \rightarrow 0} \frac{f^{\prime}(a+h)-f^{\prime}(a-h)}{h} \\ \\
& = \frac{1}{2} \lim \limits_{h \rightarrow 0}\left[\frac{f^{\prime}(a+h)-f^{\prime}(a)}{h}+\frac{f^{\prime}(a-h)-f^{\prime}(a)}{-h}\right] \\ \\
& = \frac{1}{2} \left[ f^{\prime \prime}(a) + f^{\prime \prime} (a) \right] \\ \\
& = f^{\prime \prime}(a)
\end{aligned}
$$

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